In my notes the condition number of a non-singular rectangular matrix is defined as $$K(A)=\|A\|\|A^{-1}\|$$
and the 2-norm of a rectangular matrix as: $$\|A\|_2=\sqrt{\max_i |\lambda_i(A^\top A)|}$$, with $\lambda_i(B) $ meaning i-th eigenvalue of B
Then it gives the following formula for the condition number with respect to the 2-norm: $$K_2(A)=\frac{\sqrt{\max_i|\lambda_i(A^\top A)}|}{\sqrt{\min_i|\lambda_i(A^\top A)|}}$$
I am trying to prove this formula. I've seen online that they use singular values, but I think is not necessary .
This is my proof:
Using the definitions:
$K_2(A)=\|A\|_2\|A^{-1}\|_2 = \sqrt{\max_i|\lambda_i(A^\top A)|} \sqrt{\max_i|\lambda_i((A^{-1})^\top A^{-1})|}$.
So all I have too prove is that:
$\sqrt{\max_i|\lambda_i((A^{-1})^\top A^{-1})|}=1/\sqrt{\min_i|\lambda_i(A^\top A)|}$
If $\lambda$ is an eigenvalue of a non singular square matrix C with eigenvector $v$, then: $Cv=\lambda v$ so, since C is non singular: $v=\lambda C^{-1}v$ , so $(1/\lambda)v= C^{-1}v$ , so $1/\lambda$ is an eigenvalue of $C^{-1}$. So I have proven the eigenvalues of the inverse are the reciprocal of the eigenvalues. Using that:
$\sqrt{\max_i|\lambda_i((A^{-1})^\top A^{-1})|}=\sqrt{\max_i|\lambda_i(A A^\top)^{-1})|} = 1/\sqrt{\min_i|\lambda_i(A A^\top)|}$
Now $A A^\top $ is an square matrix and it is invertible since A is invertible. So $0$ is not an eigenvalue of $A A^\top$ .(Moreover it is symmetric so it has real eigenvalues if that is useful )
, but $A^T A$ and $AA^T $ have the same non-zero eigenvalues, since $ 0$ is not an eigenvalue they have the same eigenvalues so $1/\sqrt{\min_i|\lambda_i(A A^\top)|}= 1/\sqrt{\min_i|\lambda_i(A^\top A )|}$ which is what I needed to complete the proof
What do you think, is it ok? If it is, why do people go through all the trouble of using the more complicated singular values approach ?. If it is not correct, where is it wrong?